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Dynamics: Newton's Second Law

  1. Jan 8, 2015 #1
    1. The problem statement, all variables and given/known data
    Here's the problem:
    A particular racecarcan cover a quartermile track (402m) in 6.40s starting from standstill. Assuming that the acceleration is constant, how many "g's"(all i know is that it means accleration due to gravity) does the driver experience? If the combined mass of the driver and the racecar is 485kg, what horizontal force must the road exert on the tires?

    2. Relevant equations
    1. Δy(displacement) = [(Velocityinitial)(Δt)] + ((1/2)acceleration)(Δt)2
    2. Forcegrav + (acceleration)(mass)= Forcetotal
    I'm not even sure if the second equation even exists but thats what i made out from the problem.
    3. The attempt at a solution
    I tried solving for the "g's" using the first equation and i substituted the possible things:

    402m = [(0m/s)(6.4s)] + [((1/2)a)(6.4s)]
    i don't know if this is the right equation but its the closest that i can get

    then i tried solving for the horizontal force with the second equation and substituted the given:
    [(-9.8m/s2)(485kg)] + [(19.6m/s2)(485kg)] = Forcetotal

    and in case oyu are wondering where i got the acceleration(19.6m/s2) its from the first equation i solved (which i think is wrong)

    Thank you in Advance!

  2. jcsd
  3. Jan 8, 2015 #2


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    Hi, yandereni.

    You did well in the first part. It looks correct.

    What are the directions of the two forces you've included here? What is the direction the question is concerned with?
  4. Jan 8, 2015 #3
    I looked at the problem again and there were no directions stated. And when i told my other classmates about the first part if i did it right, they said no and the answer should be 2.00m/s2.
  5. Jan 8, 2015 #4


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    what's this then, eh?
    That's wrong. The car would cover some 41m at that acceleration.
  6. Jan 8, 2015 #5


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    Maybe they meant 2g, not 2m/s^2?
    2g is about 19.62m/s^2.
  7. Jan 8, 2015 #6
    oh, now i understand. but can i solve the horizontal force without any directions given?
  8. Jan 8, 2015 #7


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    But you do know which direction the car is moving, don't you? All you need to make sure when setting up your equation is not to include forces that are not acting (i.e., have no component) in the direction you're interested in.
  9. Jan 8, 2015 #8
    Thank you very much! I got the right answer and i get the concept now. Thanks a lot! :)
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